Problem: Find $\lim_{x\to\infty}\dfrac{2-3x}{x+\sin(x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $-3$ (Choice C) C $0$ (Choice D) D The limit doesn't exist
Answer: When dealing with limits that include $\sin(x)$, it's important to remember that $\lim_{x\to\infty}\sin(x)$ doesn't exist, as $\sin(x)$ keeps oscillating between $-1$ and $1$ forever. ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${2}$ $y$ $x$ $y=\sin(x)$ This doesn't necessarily mean that our limit doesn't exist. Think what happens to $\dfrac{2-3x}{x+\sin(x)}$ as $x$ increases towards positive infinity. $\sin(x)$ oscillates between $-1$ and $1$. This can be represented mathematically by the following double inequality: $\dfrac{2-3x}{x-1}\leq\dfrac{2-3x}{x+\sin(x)}\leq\dfrac{2-3x}{x+1}$ The result is a graph that's always between the graphs of $y=\dfrac{2-3x}{x\pm 1}$ (the dashed lines). ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}2}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}2}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ $y$ $x$ Since $\lim_{x\to\infty}\dfrac{2-3x}{x\pm 1}=-3$, so must our limit be equal to $-3$. In conclusion, $\lim_{x\to\infty}\dfrac{2-3x}{x+\sin(x)}=-3$.